Mathematics 101 Online
OpenStudy (anonymous):

Find the equation of the tangent to the graph at the indicated point. HINT [Compute the derivative algebraically] f(x) = x^2 − 1; a = 1 I found the slope using lim as h approaches 0, f(1+h) - f(1) / h and got lost during that part.

OpenStudy (anonymous):

$f'(x) = \lim_{h \to 0} \frac{f(x+h)- f(x)}{h}$ for x = 1: $\lim_{h \to 0} \frac{f(1+h)- f(1)}{h} \lim_{h \to 0} \frac{(1+h)^2 -1 - (1^2 - 1)}{h} = \lim_{h \to 0} \frac{1+2h + h^2 -1 - 1 + 1)}{h} =$ $\lim_{h \to 0} \frac{2h+h^2}{h} = \lim_{h \to 0} \frac{h(2+h)}{h} = \lim_{h \to 0} \,\,(2+h) = 2$

OpenStudy (anonymous):

$y-y_1 = m(x - x_1)$ $y_1 = f(1) = 1^2 - 1= 0$ equation of the tangent line at that point therefore is $y = 2(x-1)$ $y = 2x-2$

OpenStudy (anonymous):

That's what I ended up getting for the second part. Thank you!

OpenStudy (anonymous):

you're welcome